Question
How many grams of CH3COOH are needed to make 4.2 liters of aqueous solution of pH=3.4?Be sure to back correct from equilibrium molecular acid molarity to starting molecular acid molarity.
Answers
I can tell you how to solve the problem BUT I have no idea what your second sentence says.
CH3COOH is a weak acid, therefore, one must determine CH3COOH from the Ka and the known H3O^+.
CH3COOH + H2O ==> H3O^+ + CH3COO^-
Ka = (H3O^+)(CH3COO^-)/(CH3COOH)
If pH = 3.4, use pH = -log(H3O^+) to solve for (H3O^+). Plug that into the Ka expression for (H3O^+) and for (CH3COO^-)
For (CH3COOH), plug in X-(H3O^+) and solve for X. That will give you the molarity of CH3COOH you need for that pH.
You want 4.2L so moles CH3COOH needed will be M x L.
moles = grams/molar mass. Solve for grams.
CH3COOH is a weak acid, therefore, one must determine CH3COOH from the Ka and the known H3O^+.
CH3COOH + H2O ==> H3O^+ + CH3COO^-
Ka = (H3O^+)(CH3COO^-)/(CH3COOH)
If pH = 3.4, use pH = -log(H3O^+) to solve for (H3O^+). Plug that into the Ka expression for (H3O^+) and for (CH3COO^-)
For (CH3COOH), plug in X-(H3O^+) and solve for X. That will give you the molarity of CH3COOH you need for that pH.
You want 4.2L so moles CH3COOH needed will be M x L.
moles = grams/molar mass. Solve for grams.
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